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(C) Let the principal sum be $P$.Given: Compound Interest $(CI)$ = $Rs. 101.50$,Rate $(r)$ = $3 \%$,Time $(t)$ = $2$ $years$.The formula for compound

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$100.00$

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(C) Let the principal sum be $P$.Given: Compound Interest $(CI)$ = $Rs. 101.50$,Rate $(r)$ = $3 \%$,Time $(t)$ = $2$ $years$.The formula for compound interest is $CI = P \left[ \left( 1 + \frac{r}{100} ight)^t - 1 ight]$.Substituting the values: $101.50 = P \left[ \left( 1 + \frac{3}{100} ight)^2 - 1 ight]$.$101.50 = P \left[ (1.03)^2 - 1 ight] = P [1.0609 - 1] = P(0.0609)$.$P = \frac{101.50}{0.0609} = \frac{1015000}{609} = Rs. \frac{5000}{3}$.Now,calculate Simple Interest $(SI)$ using $SI = \frac{P 	imes r 	imes t}{100}$.$SI = \frac{(5000/3) 	imes 3 	imes 2}{100} = \frac{5000 	imes 2}{100} = Rs. 100$.

If the compound interest on a certain sum for $2$ $years$ at $3 \%$ per annum is $Rs. 101.50$,then the simple interest (in $Rs.$) on the same sum at the same rate and for the same time will be:

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